2014 Leaving Cert Higher LevelOfficial Sample Paper 2

Section A Concepts and Skills 150 marks

Question 1 (25 marks)The random variable X has a discrete distribution. The probability that is takes a value other than13, 14, 15 or 16 is negligible.

(a) Complete the probability table below and hence calculate E(X), the expected value of X .

x 13 14 15 16P(X = x) 0.383 0.575 0.038 0.004

Note that the sum of the probabilities must be 1. Therefore P(15) = 1− 0.383− 0.575−0.004 = 0.038.E(X) = 13(0.383)+14(0.575)+15(0.038)+16(0.004) = 13.663.

(b) If X is the age, in complete years, on 1 January 2013 of a student selected at random fromamong all second-year students in Irish schools, explain what E(X) represents.

E(X) represents the mean of the ages of all second-year students in Irish schools on 1January 2013.

(c) If ten students are selected at random from this population, find the probability that exactlysix of them were 14 years old on 1 January 2013. Give your answer correct to three signifi-cant figures.

We have 10 Bernoulli trials with p = P(X = 14) = 0.575. So the probability of exactly sixsuccesses is given by(

106

)p6(1− p)4 =

10!6!4!

(0.575)6(1−0.575)4

= 210(0.575)6(0.425)4

= 0.248

correct to three significant places.

Question 2 (25 marks)

(a) Explain what is meant by stratified sampling and cluster sampling. Your explanation shouldinclude:

• a clear indication of the difference between the two methods

• one reason why each method might be chosen instead of simple random sampling.

In stratified sampling the population is divided into strata and a sample is taken by applyingrandom sampling within each of the strata. We usually ensure that the size of the samplefrom each strata is proportional to the size of the strata relative to the overall population.One reason why this method might be chosen is to ensure that each strata of the populationis represented proportionately in the overall sample.In cluster sampling the population is divided in groups and then we choose a simple randomsample of the groups. Then we create our sample by choosing simple random samples onlyfrom the selected groups (sometimes we might create the sample by choosing all of theselected groups). One reason why this method might be chosen instead of a simple randomsample is that it can reduce the cost of sampling, since we restrict our sample to only theselected groups rather than the entire population.

(b) A survey is being conducted of voters’ opinions on several different issues.

(i) What is the overall margin of error of the survey, at 95% confidence, if it is based on asimple random sample of 1111 voters.

The margin of error is1√

1111= 0.03000

correct to 4 significant places, or 3.000%.

(ii) A political party had claimed that it has the support of 23% of the electorate. Of thevoters in the sample above, 234 stated that they support the party. Is this sufficientevidence to reject the party’s claim, at the 5% level of significance?

H0: The party has the support of 23% of the electorate, or p = 0.23.The standard error of the sampling distribution, assuming that H0 is true, is given by

s =

√p(1− p)

n=

√0.23(0.77)

1111= 0.01263

correct to four significant places. Now the observed sample proportion is

p̂ =234

1111= 0.2106

The z-score corresponding to 0.2106 is

p̂− ps

=0.2106−0.23

0.01263=−1.5360

At the 5% level of significance, we reject the null hypothesis if this z-score lies outside theinterval [−1.96,1.96]. However, the calculated z-score lies inside this interval. This meansthat we do not have enough evidence at the 5% level of significance to reject the party’sclaim.

Question 3 (25 marks)

(a) Show that, for all k ∈R, the point P(4k−2,3k+1) lies onthe line l1 : 3x−4y+10 = 0.

If (x,y) = (4k−2,3k+1) then

3x−4y+10 = 3(4k−2)−4(3k+1)+10= 12k−6−12k−4+10= 0

So the equation of l1 is satisfied. Therefore(4k−2,3k+1) lies on l1.

P

Q

l1

(b) The line l2 passes through P and is perpendicular to l1. Find the equation of l2 in terms of k.

We have

3x−4y+10 = 0⇔

−4y = −3x−10⇔

y =34

x+52

Therefore the slope of l1 is 34 . Therefore the slope of l2 is 1

34=−4

3 . So l2 has slope −43 and

passes through (4k−2,3k+1). So it has equation

y− (3k+1) =−43(x− (4k−2))

or3y−3(3k+1) =−4(x− (4k−2)).

Rearranging this gives4x+3y−25k+5 = 0.

(c) Find the value of k for which l2 passes through the point Q(3,11).

The equation of l2 is4x+3y−25k+5 = 0.

Now (3,11) lies on l2 if and only if 4(3)+ 3(11)− 25k+ 5 = 0⇔ 25k = 50⇔ k = 2. Sothe k = 2 is the required value.

(d) Hence, or otherwise, find the co-ordinates of the foot of the perpendicular from Q to l1.

When k = 2 the equation of l2 is

4x+3y−45 = 0.

So to find the required point, we solve

3x−4y+10 = 04x+3y−45 = 0

simultaneously.This is equivalent to

12x−16y+40 = 012x+9y−135 = 0

Subtracting yields−25y+175 = 0.

Therefore 25y = 175 and y = 17525 = 7.

Now 3x−4(7)+10 = 0⇔ 3x = 4(7)−10 = 18⇔ x = 6. So the foot of the perpendicularfrom Q to l1 has co-ordinates (6,7).

Question 4 (25 marks)The centre of a circle lies on the line x+2y−6 = 0. The x-axis and the y-axis are tangents to thecircle. There are two circles that satisfy these conditions. Find their equations.

Consider the diagram below:

(b,b)

(−a,a)

(d,−d)

(−c,−c)

a

a

c

c

b

b

d

d

Note that this diagram is not meant to represent the solution. It is just meant to illustrate thepossibilities for circles that are tangent to both axes. In the diagram, a, b, c and d are theradii of the circles. We can see from this diagram that if (x,y) is the centre of a circle thathas both the x-axis and the y-axis as tangents, then either

• Case 1: y = x

• Case 2: y =−x

In either case the radius is either x or −x. Since the radius is positive, we can say that theradius is |x| in either case. . .

Case 1: y = x. We are also told that x+2y−6 = 0. Subsituting x for y in the latter equationgives

x+2x−6 = 0⇔ 3x−6 = 0⇔ x = 2.

Now y = x so y = 2. Therefore the centre of the circle has co-ordinates (2,2) and the radiusis 2. Therefore in this case the circle has equation

(x−2)2 +(y−2)2 = 4.

Case 2: y =−x. As before we use this to substitute −x for y in the equation x+2y−6 = 0.This gives

x+2(−x)−6 = 0⇔−x−6 = 0⇔ x =−6.

It follows that y = −(−6) = 6. So in this case the centre has co-ordinates (−6,6) and theradius is 6. So this circle has equation

(x+6)2 +(y−6)2 = 36.

Question 5 (25 marks)The diagram below shows the graph of the function f : x 7→ sin2x. The line 2y = 1 is also shown.

The graph of f is in blue. The graph of g is in green and the graph of h is in red.

(a) On the same diagram above sketch the graphs of g : x 7→ sinx and h : x 7→ 3sin2x. Indicateclearly which is g and which is h.

(b) Find the co-ordinates of the point P in the diagram.

P is a point of intersection of the line y = 12 and the curve y = sin2x. So the x-co-ordinate

of P is a solution of the equation sin2x = 12 .

Therefore either2x =

π

6+2nπ,n ∈ Z

or2x =

5π

6+2nπ,n ∈ Z.

In the first case, we getx =

π

12+nπ,n ∈ Z

and in the second case we get

x =5π

12+nπ,n ∈ Z.

So x is one of the following numbers

. . . ,π

12,5π

12,13π

12,17π

12,25π

12,29π

12, . . .

From the diagram we see that5π

4≤ x≤ 3π

2so the only possibility is that

x =17π

12.

Clearly the y-co-ordinate of P is 12 since it lies on the line 2y = 1. So the co-ordinates of P

are(17π

12 , 12

).

Question 6A (25 marks)Explain, with the aid of an example, what is meant by proof by contradiction.Note: you do not need to provide the full proof involved in your example. Give sufficient outlineto illustrate how contradiction is used.

Explanation:

To prove a statement by contradiction, we assume that the statement is false and then provethat this assumption contradicts another statement that is known to be true.

Example:

Consider the statement

The lines y = x and y = x+1 do not intersect.

We can prove this by contradiction as follows.Assume that that the statement is false. So there is some point (a,b) that lies on both lines.Therefore b = a since the point is on the line y = x and b = a+ 1 since the point is on theline y = x+1.Combining these equation, we get

a = a+1

which implies that0 = 1.

However this contradicts that fact that 0 6= 1. Therefore our original assumption is false.So we have proved that the lines y = x and y = x+1 do not intersect.

Question 6B (25 marks)ABC is a triangle.

D is the point on BC such that AD ⊥ BC. E isthe point on BC such that BE ⊥ AC.

AD and BE intersect at O.

Prove that |∠DOC|= |∠DEC|.

A

B

C

D

E

O

Consider the quadrilateral DOEC. We have

|∠CDO|+ |∠OEC|= 90◦+90◦ = 180◦.

Therefore DOEC is a cyclic quadrilateral (by the converse of Corollary 5).Therefore ∠DOC and ∠DEC are angles standing on the same arc of a circle (the circumcir-cle of DOEC).Therefore, by Theorem 19

|∠DOC|= |∠DEC|

as required.

Section B Contexts and Applications 150 marks

Question 7 (75 marks)The King of the Hill triathlon race in Kinsale consists ofa 750 metre swim, followed by a 20 kilometre cycle, fol-lowed by a 5 kilometre run.

The questions below are based on the data from the 224athletes who completed this triathlon in 2010.

Máire is analysing data from the race, using statisticalsoftware. She has a data file with each competitor’s timefor each part of the race, along with various other detailsof the competitors.Máire gets the software to produce some summary statistics and it produces the following table.Three of the entries in the table have been removed and replaced with question marks (?).

Swim Cycle RunMean 18.329 41.927 ?Median 17.900 41.306 ?Mode #N/A #N/A #N/AStandard Deviation ? 4553 3409Sample Variance 10.017 20.729 11.622Skewness 1.094 0.717 0.463Range 19.226 27.282 20.870Minimum 11.350 31.566 16.466Maximum 30.576 58.847 37.336Count 224 224 224

Máire produces histograms of the times for the three events. Here are the histograms, withouttheir titles.

(a) (i) Use the summary statistics in the table to decide which histogram corresponds to eachevent. Write the answers above the histograms.

We can use the minimum and maximum values to match the histograms to the events.The leftmost histogram has a minimum between 16 and 18. From the table we see that theonly event that matches this is the run.The rightmost histogram has a minimum value somewhere between 10 and 12, so that mustbe the swim.Therefore the middle histogram must be the cycle, and indeed it matches the maximum andminimum data for the cycle.

(ii) The mean and the median time for the run are approximately equal. Estimate this valuefrom the corresponding histogram.

mean ≈median ≈ 25 minutes.

From the histogram it seems that about half of the data lies to the left of 25, so that is ourestimate for the median.

(iii) Estimate from the relevant histogram the standard deviation of the times for the swim.

standard deviation ≈ 3 minutes

The empirical rule says that 95% percent of the data lies within 2 standard deviations of themean.A reasonable guess from the histogram is that 95% of the data lies in between 12 minutesand 24minutes, so we might guess that the standard deviation is about 24−12

4 = 3.

(iv) When calculating the summary statistics, the software failed to find a mode for the datasets. Why do you think this is?

If no mode was found, it must be that there were no identical finish times recorded perevent. This is not surprising as the times are probably recorded to the nearest second oreven hundredth of a second, and any value within the range is possible. So there are a verylarge number of possible values, and it is unlikely to have two finish times that are exactlythe same. If every finish time recorded is unique, then the software will fail to find a mode.

Máire is interested in the relationships between the athletes’ performance in the three differentevents. She produces the following three scatter diagrams.

(b) Give a brief summary of the relationship between performance in the different events, basedon the scatter diagrams.

There is a positive correlation between the cycle times and the swim times.There is a positive correlation between the run times and the swim times.There is a strong positive correlation between the run times and the cycle times.

(c) The best-fit line for the run-time based on swim-time is y = 0.53x+ 15.2. The best-fit linefor run-time based on cycle-time is y = 0.58x+ 0.71. Brian did the swim in 17.6 minutesand the cycle in 35.7 minutes. Give your best estimate of Brian’s time for the run, and justifyyour answer.

Based on his swim-time we estimate his run-time to be

0.53(17.6)+15.2 = 24.528.

Based on his cycle-time we estimate his run-time to be

0.58(35.7)+0.71 = 21.416.

It seems reasonable to take the mean of these two estimates as our best guess. Therefore weestimate his run-time to be

24.528+21.4162

= 22.972 minutes.

The mean finishing time for the overall event was 88.1 minutes and the standard deviation was10.3 minutes.

(d) Based on an assumption that the distribution of overall finishing times is approximatelynormal, use the empirical rule to complete the following sentence:

”95% of the athletes took between 67.5 and 108.7 minutes to complete

the race.”

The empirical rule says that 95% of the data lies within two standard deviations of the mean.In this case that means that 95% of the data lies between 88.1−2(10.3) and 88.1+2(10.3).

(e) Using the normal distribution tables, estimate the number of athletes who completed the racein less than 100 minutes.

Given a mean of 88.1 minutes and a standard deviation of 10.3 minutes, the standardisedz-score corresponding to 100 minutes is

100−88.110.3

= 1.155

correct to three decimal places. Now using the tables we see that P(z ≤ 1.155) = 0.8760correct to four decimal places. So we estimate that the number of athletes who completedthe race in less than 100 minutes is 0.8760(224) ≈ 196. So we estimate that about 196athletes completed the race in less than 100 minutes.

(f) After the event, a reporter wants to interview two people who took more than 100 minutesto complete the race. She approaches athletes at random and asks them their finishing time.She keeps asking until she finds someone who took more than 100 minutes, interviews thatperson, and continues until she finds a second such person. Assuming the athletes are coop-erative and truthful, what is the probability that the second person that she interviews will bethe sixth person she approaches?

From part (e), we know that the probability of a randomly selected athlete having a finishtime of more than 100 minutes is 1−P(less than 100 minutes) = 1−0.876 = 0.124.So we model this problem as a sequence of Bernoulli trials where the probability of successin each trial is

p = 0.124.

The probability that the second success occurs on the sixth trial is

P(exactly one success in the first 5)×P(success in trial 6)

Now

P(exactly one in first 5) = 5× p(1− p)4

= 5(0.124)(0.876)4

= 0.365

correct to three decimal places.Also,

P(success in trial 6) = p = 0.124.

So the answer is0.365×0.124 = 0.045

correct to three decimal places.

Question 8 (50 marks)A stand is being used to prop up a portable solar panel. It consists ofa support that is hinged to the panel near the top, and an adjustablestrap joining the panel to the support near the bottom.

By adjusting the length of the strap, the angle between the panel andthe ground can be changed.

(a) Find the length of the strap |DE| such that the angle α between the panel and the ground is60◦.

Consider the diagram below in which we have marked some of the lengths. Note that |DC|must be 20cm since we are told that |AB|= 30cm. Similarly |CE|= 18cm since we are toldthat |CF |= 22cm.

20cm

5cm

5cm

18cm

4cm

A F

B

C

D

E

α

Now suppose that α = 60◦. Then we apply the Sine Rule to the triangle4ACF . So

sin60◦

22=

sin(∠F)

25

Therefore

sin(∠F) =25sin(60)

22= 0.9841

correct to four decimal places. Therefore |∠F |= sin−1(0.9841) = 79.77◦. Now we use thisto calculate ∠C. So

|∠C|= 180−60−79.77 = 40.23◦.

Now we apply the Cosine Rule to the triangle4CDE. Thus,

|DE|2 = 202 +182−2(20)(18)cos(40.23◦)= 400+324−549.7= 174.3

Therefore|DE|=

√174.3 = 13.20cm.

So the length of the strap when α = 60◦ is 13.20cm.

(b) Find the maximum possible value of α , correct to the nearest degree.

The maximum possible value of α will occur when the stand is set so that CF is vertical.In that case 4ACF is a right angled triangle with the hypotenuse |AC| = 25cm. The sideopposite the angle α is |CF |= 22cm. Therefore in this case,

sinα =2225

= 0.88.

Soα = sin−1(0.88) = 61.64 = 62◦

correct to the nearest degree.

Question 9 (25 marks)A regular tetrahedron has four faces, each ofwhich is an equilateral triangle.

A wooden puzzle consists of several pieces thatcan be assembled to make a regular tetrahedron.The manufacturer wants to package the assem-bled tetrahedron in a clear cylindrical container,with one face flat against the bottom.

If the length of one edge of the tetrahedron is 2a,show that the volume of the smallest possible

cylindrical container is

(8√

69

)πa3.

Consider the base of the cylindrical container together with the base of the tetrahedron drawnin the diagram below:

O

B

A

2a

C

Now O is the circumcentre of an equilateral triangle. So

|∠BOA|= |∠AOC|= |∠COB|

and since these angles sum to 360◦, we must have |∠BOA|= 120◦.Now consider the triangle4AOB. It is an isosceles triangle, so |∠ABO|= |∠BAO|. Since

|∠ABO|+ |∠BAO|+ |∠BOA|= 180◦

we get |∠ABO| = 30◦. Now we apply the Sine Rule to the triangle 4AOB to get |OA|sin30◦ =

2asin120◦ or

|OA|= 2asin30◦

sin120◦= 2a

12√3

2

=2a√

3.

So the radius of the cylinder is 2a√3.

Solution continued on next page . . .

Now let D be the top point of the tetrahedron and drop a vertical line from D to O to createa right angled triangle as shown below. Let h be the height of the cylinder.

2a√3

2a

h

D

O A

Therefore, by Pythagoras’ Theorem,

h2 +

(2a√

3

)2

= (2a)2

which implies that h2 = 4a2− 4a2

3 = 8a2

3 . Therefore

h = a

√83= 2a

√23.

Now the volume of a cylinder with height 2a√

23 and radius 2a√

3is

π

(2a√

3

)2

2a

√23= π

8a3√

23√

3.

Now multiply this last expression above and below by√

3 to obtain a volume of

π8a3√

69

as required.